 # What is a tight lower bound to sum^n_(i=1)(1)/(a+x_i) under the restrictions sum^n_(i=1)x_i=0 and sum^n_(i=1)x^2_i=a^2 ? Mehlqv 2022-08-16 Answered
What is a tight lower bound to $\sum _{i=1}^{n}\frac{1}{a+{x}_{i}}$ under the restrictions $\sum _{i=1}^{n}{x}_{i}=0$ and $\sum _{i=1}^{n}{x}_{i}^{2}={a}^{2}$ ?
Conjecture: due to the steeper rise of $\frac{1}{a+x}$ for negative $x$, one may keep those values as small as possible. So take $n-1$ values ${x}_{i}=-q$ and ${x}_{n}=\left(n-1\right)q$ to compensate for the first condition. The second one then gives ${a}^{2}=\sum _{i=1}^{n}{x}_{i}^{2}={q}^{2}\left(\left(n-1{\right)}^{2}+n-1\right)={q}^{2}n\left(n-1\right)$. Hence,
$\sum _{i=1}^{n}\frac{1}{a+{x}_{i}}\ge \frac{n-1}{a\left(1-1/\sqrt{n\left(n-1\right)}\right)}+\frac{1}{a\left(1+\left(n-1\right)/\sqrt{n\left(n-1\right)}\right)}$
should be the tight lower bound.
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Yes, this is indeed the minimum value (assuming $a>0$).

Denote $K=\left\{x\in {\mathbb{R}}^{n}\mid \sum _{k=1}^{n}{x}_{k}=0,\sum _{k=1}^{n}{x}_{k}^{2}={a}^{2}\right\}$ and let $f\left(x\right)=\sum _{k=1}^{n}\left(a+{x}_{k}{\right)}^{-1}$ attain its minimum at $x=\overline{x}\in K$ (it does so, as a continuous function on $\left\{x\in K\mid f\left(x\right)⩽f\left(0\right)\right\}$ which is compact). Then, by Lagrange multiplier theorem, there are ${\lambda }_{1},{\lambda }_{2}\in \mathbb{R}$ such that for each k we have $\left(a+{\overline{x}}_{k}{\right)}^{-2}={\lambda }_{1}+{\lambda }_{2}{\overline{x}}_{k}$. Then the positive numbers ${y}_{k}=a+{\overline{x}}_{k}$ are solutions of
${y}^{2}\left({\lambda }_{1}-{\lambda }_{2}a+{\lambda }_{2}y\right)=1.$
But this equation has at most two positive solutions. Thus, at most two values among ${\overline{x}}_{k}$ are distinct, and in fact exactly two. So, let $m$ values of ${\overline{x}}_{k}$ equal $b>0$, where $0, and the remaining $n-m$ values equal $c<0$. We get a system for $b$ and $c$, solve it, and finally obtain
$af\left(\overline{x}\right)=n+{\left(1-\frac{1}{n}+\frac{n-2m}{\sqrt{nm\left(n-m\right)}}\right)}^{-1}.$
The least possible value of this is at $m=1$.

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